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Is there a nice short version of the L squared operator

  1. Oct 21, 2007 #1
    This is a little bit of a weird one.

    The question asks me to prove that [tex][\hat{L}^2,\hat{L}_z ][/tex] is equal to zero. I know what to do with it and how commutators work, and I know that L2 = Lx2 + Ly2 + Lz2 (where the 2 means squared) and I have expressions for Lx2, Ly2 and Lz2, but I was just wondering if theres a nice compact way of writing L2 out, as the version I'll get after adding the 3 expressions is really long and is a ridiculous amount of (albeit simple) working out, in which I'm likely to make sign errors (which I did on the previous question).

    If I have to go this long winded way so be it (and I will do - just like to have something to aim at), but if it could be simplfied then I'm sure it'd be a slightly easier question.

    Thanks guys
    Brewer
     
  2. jcsd
  3. Oct 21, 2007 #2

    Gokul43201

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    There are no long expressions that are required for evaluating this commutator. You can presumably use the elementary commutation relations [itex][L_x,L_y]=i\hbar L_z~[/itex] etc.
     
  4. Oct 22, 2007 #3
    Yes, that was the first part of the question to come up with that, but I'm not sure how you go about doing it like that. My shabby notes don't seem to be much help - they just tell me that L2 = Lx2 + Ly2 + Lz2.
     
  5. Oct 22, 2007 #4

    cristo

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    Well, like Gokul says, use the commutation relations. You know the form for l^2, so write out the commutator. Presumably you can then expand this and simplify using the rules of commutators that you know.

    Why don't you have a go, and post what you get. Then someone will be able to point you in the right direction.
     
  6. Oct 22, 2007 #5

    CompuChip

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    And you'll probably want to use something like
    [AB, C] = A[B, C] + [B, C]A
    (easily proved, just write it out)
     
  7. Oct 22, 2007 #6

    Gokul43201

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    I think compuchip may have been typing in a bit of a hurry. That should read: [AB,C] = A[B,C] + [A,C]B
     
  8. Oct 22, 2007 #7

    CompuChip

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    Hehe, obviously. My apologies.
     
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